Problem: Divide the following complex numbers. $ \dfrac{-4-2i}{1-i}$
Solution: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${1+i}$ $ \dfrac{-4-2i}{1-i} = \dfrac{-4-2i}{1-i} \cdot \dfrac{{1+i}}{{1+i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(-4-2i) \cdot (1+i)} {(1-i) \cdot (1+i)} = \dfrac{(-4-2i) \cdot (1+i)} {1^2 - (-1i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(-4-2i) \cdot (1+i)} {(1)^2 - (-1i)^2} = $ $ \dfrac{(-4-2i) \cdot (1+i)} {1 + 1} = $ $ \dfrac{(-4-2i) \cdot (1+i)} {2} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({-4-2i}) \cdot ({1+i})} {2} = $ $ \dfrac{{-4} \cdot {1} + {-2} \cdot {1 i} + {-4} \cdot {1 i} + {-2} \cdot {1 i^2}} {2} $ Evaluate each product of two numbers. $ \dfrac{-4 - 2i - 4i - 2 i^2} {2} $ Finally, simplify the fraction. $ \dfrac{-4 - 2i - 4i + 2} {2} = \dfrac{-2 - 6i} {2} = -1-3i $